Circular orbits arise whenever the gravitational force on a satellite equals the centripetal force needed to move it with uniform circular motion.
|Fc = Fg||⇒||m2v2||=||Gm1m2||⇒||v2 =||Gm1|
Substitute this expression into the formula for kinetic energy.
|K =||1||m2v2 =||1||m2||⎛
Note how similar this new formula is to the gravitational potential energy formula.
|2||r||K = −||1||Ug|
The kinetic energy of a satellite in a circular orbit is half its gravitational energy and is positive instead of negative. When U and K are combined, their total is half the gravitational potential energy.
|E = K + Ug = −||1||Ug + Ug =||1||Ug|
|E = −||Gm1m2|
The gravitational field of a planet or star is like a well. The kinetic energy of a satellite in orbit or a person on the surface sets the limit as to how high they can "climb out of the pit". A satellite in a circular orbit is halfway out of the pit (or halfway in, for you pessimists).
Start by determining the radius of a geosynchronous orbit. There are several ways to do this (which includes looking it up somewhere), but the traditional way is to start from the principle that the centripetal force on a satellite in a circular orbit is provided by the gravitational force of the earth on the satellite. Combine this with the formula for the speed of an object in uniform circular motion. The algebra is somewhat tedious and has been condensed in the derivation below.
|Fc =||mv2||=||Gm1m2||= Fg||&||v =||Δs||=||2πr||⇒||r =||⎛
|(6.67 × 10−11 Nm2/kg2)(5.98 × 1024 kg)(24 × 60 × 60 s)2||⎞⅓
rf = 4.225 × 107 m (geostationary orbit)
Next, use the relationship derived in the previous problem to determine the total energy of the satellite in orbit. This will be the final energy of the system.
|Ef = Kf + Uf =||Uf|
|Ef = −||Gm1m2|
|Ef = −||(6.67 × 10−11 Nm2/kg2)(5.98 × 1024 kg)(5000 kg)|
|2(4.225 × 107 m)|
Ef = −2.360 × 1010 J = −23.60 GJ (geostationary orbit)
To satisfy the minimum energy requirements of this problem the satellite should be launched from someplace on the equator where the speed of rotation (and thus the kinetic energy) is a maximum.
|vi =||Δs||=||2πr||=||2π(6.37 × 106 m)||= 463.2 m/s (on the equator)|
|Δt||T||(24 × 60 × 60 s)|
The initial energy of the satellite is the gravitational potential energy it has on the earth's surface plus the kinetic energy it has due to the earth's rotation. (Remember, gravitational potential energy is negative.)
Ei = Ki + Ui
|Ei =||1||mvi2 −||Gm1m2|
|Ei =||1||(463.2 m/s)(5000 kg)2 −||(6.67 × 10−11 Nm2/kg2)(5.98 × 1024 kg)(5000 kg)|
|2||(6.37 × 106 m)|
Ei = −3.125 × 1011 J = −312.5 GJ (on the equator)
Subtract the initial and final energies to finish the problem. State the aswer with an appropriate number of significant digits (two, since the period of the earth's rotation — 24 h — is only accurate to two significant digits).
ΔE = Ef − Ei = (-23.60 GJ) − (-312.5 GJ) = 290 GJ
The total energy of a satellite is just the sum of its gravitational potential and kinetic energies. Assuming that energy is conserved (which it is for the most part in the vacuum of space), the total energy of the satellite would remain constant. Set the total energy of the satellite at apogee equal to the total energy at perigee and solve for va.
|va2 − vp2 = 2GM||⎛
|vp2 + 2GM||⎛
For the same reason that mechanical energy is conserved angular momentum is also conserved as the satellite moves from perigee to apogee. The speed at apogee can be determined quite simply from this principle. Set the angular momentum of the satellite at apogee equal to the angular momentum at perigee and solve for va.
|La = Lp||⇒||mvara = mvprp||⇒||va =||rp||vp|
|Fc = Fg||⇒||mv2||=||GMm||⇒||v = √||GM|
For the second and third parts of this problem, apply the impulse-momentum theorem …
J = Δp = mΔv = m(v − vp)
… and the work-energy theorem.
|W = ΔK =||1||mv2 −||1||mvp2 =||1||m||⎛